The position $$x$$ of a particle w.r.t. time $$t$$ along $$x$$-axis is given by $$x = 9{t^2} - {t^3},$$   where $$x$$ is in metre and $$t$$ in $$sec.$$  What will be the position of this particle when it achieves maximum speed along the $$+ x$$  direction?

Solution :
Given, the position $$x$$ of a particle w.r.t. time $$t$$ along $$x$$-axis
$$x = 9{t^2} - {t^3}\,......\left( {\text{i}} \right)$$
Differentiating Eq. (i), w.r.t. time, we get speed, i.e.
$$\eqalign{ & v = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {9{t^2} - {t^3}} \right) \cr & {\text{or}}\,v = 18t - 3{t^2}\,......\left( {{\text{ii}}} \right) \cr} $$
Again differentiating Eq. (ii), with respect to time, we get acceleration, i.e.
$$\eqalign{ & a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left( {18t - 3{t^2}} \right) \cr & {\text{or,}}\,a = 18 - 6t\,......\left( {{\text{iii}}} \right) \cr} $$
Now, when speed of particle is maximum, its acceleration is zero, i.e.
$$\eqalign{ & a = 0 \cr & {\text{i}}{\text{.e}}{\text{.}}\,18 - 6t = 0 \cr & {\text{or}}\,t = 3\,s \cr} $$
Putting in Eq. (i), we obtain position of particle at the time
$$\eqalign{ & x = 9{\left( 3 \right)^2} - {\left( 3 \right)^3} \cr & = 9\left( 9 \right) - 27 \cr & = 81 - 27 \cr & = 54\,m \cr} $$